QJEEAdv16Ph3344: JEE Advanced 2016 Physics - Class 12 JEEAdv16Ph3344 Solution: A transparent slab of thickness d has a refractive index n (z)…

From Snell's law,

$$\begin{array}{rl}{n}_1\sin {\theta }_i=n\left(z_1\right)\sin {\theta }_{z_1}=n\left(z_2\right)\sin {\theta }_{z_2}& =\dots \dots \\ \\ \dots & =n(d)\sin {\theta }_d=n_2\sin {\theta }_f\end{array}$$

The lateral displacement of a ray by a parallel slab depends on the angle of incidence $\theta_i$ , refractive index $n_1$ , slab thickness $d$ and the refractive index of the slab $n$ (it is independent of $n_2$ because $l$ is measured before the start of medium 2, there is no effect of $n_2$ on $l$ ). Consider the parallel slab shown in the figure.


By Snell's law, $n_1 \sin i_1=n \sin r_1$ and $n \sin i_2=$ $n_2 \sin r_2$ . By geometry, $i_2=r_1$ and the lateral displacement is given by

$$l=d\frac{\sin r_1}{\cos r_1}=\frac{d{n}_1\sin i_1}{\sqrt{n^2-n_1^2{\sin }^2{i}_1}}.$$

Hence, $l$ is independent of $n_2$ but it is dependent of $n(z)$ .